Forces And Newtons Laws Of Motion

the single point at which the entire weight of the body appears to act; for a uniform symmetric body this is the geometric centre

weight acting vertically downwards from the centre of gravity; tension in the cable acting vertically upwards on the lamp

an object continues at rest or moves with constant velocity in a straight line; unless acted on by a resultant external force

$$a=\frac{F}{m}=\frac{3600}{1200}$$ $$a=3.0{\text{ m s}}^{-2}$$

acceleration is directly proportional to resultant force when mass is constant; therefore the acceleration also doubles to 6.0 m s⁻²

initially weight is greater than drag, so there is a downward resultant force and the skydiver accelerates downwards; as speed increases the drag force increases; eventually drag becomes equal in magnitude to weight; the resultant force is then zero, so the acceleration is zero and the skydiver falls at constant velocity

the gravitational force exerted per unit mass; on a small test mass placed at that point

$$m=\frac{W}{g}=\frac{96}{1.6}$$ $$m=60\text{ kg}$$

mass is a measure of the quantity of matter, which does not change with location; weight is the gravitational force on the object, given by W = mg; the gravitational field strength g is smaller in orbit than at Earth's surface, so the weight is smaller

it is small and dense, so air resistance is negligible compared with weight; it is ferromagnetic, so it can be held and released by the electromagnet

Starting from the equation of motion: $$s=\frac{1}{2}g{t}^{2}$$ Rearranging for g: $$g=\frac{2s}{{t}^{2}}=\frac{2\times 0.840}{{0.415}^{2}}$$ $$g=9.75{\text{ m s}}^{-2}$$

a graph uses every data point so random errors in individual readings are averaged out; the gradient of the line of best fit gives g directly; any systematic error producing a non-zero intercept is identifiable on the graph

a delay between switching off the current and the ball releasing from the electromagnet; this makes t slightly too long, which gives a calculated g that is smaller than the true value

when body A exerts a force on body B; body B exerts a force on body A that is equal in magnitude and opposite in direction

the swimmer pushes water backwards; the water pushes the swimmer forwards with an equal and opposite force; the two forces are equal in magnitude / act in opposite directions / are the same type of force (contact) / act on two different bodies (any two)

The true weight of the passenger: $$W=mg=75\times 9.81$$ $$W=736\text{ N}$$ acting vertically downwards on the passenger. During free fall the only force acting on the passenger is their weight, so the resultant force is 736 N downwards. Using ΣF = ma, the acceleration is: $$a=\frac{736}{75}=9.81{\text{ m s}}^{-2}$$ downwards, which equals g. The scale and passenger now accelerate downwards together at the same rate, so there is no contact force between them; the scale therefore reads zero, often described as “apparent weightlessness”. This is consistent with Newton’s second law because no contact force is needed to provide the passenger’s downward acceleration. A Newton’s third law pair acting on the passenger is the gravitational pull of the Earth on the passenger (downwards) and the gravitational pull of the passenger on the Earth (upwards), which are equal in magnitude, opposite in direction, of the same type, and act on different bodies.